A torque of Tp = 1320 N-m is applied to gear D of the gear train shown. The bearings shown allow the shafts to rotate freely. Assume shafts (1) and (2) are solid 35-mm-diameter steel shafts. Determine the magnitude of the maximum shear stress acting in shaft (2). 230-mm dia. T Tp B D (1) 150-mm dia. O 171.6 MPa O 96.0 MPa O 156.8 MPa O 184.8 MPa

A torque of Tp = 1320 N-m is applied to gear D of the gear train shown. The bearings shown allow the shafts to rotate freely. Assume shafts (1) and (2) are solid 35-mm-diameter steel shafts. Determine the magnitude of the maximum shear stress acting in shaft (2). 230-mm dia. T Tp B D (1) 150-mm dia. O 171.6 MPa O 96.0 MPa O 156.8 MPa O 184.8 MPa

Principles of Foundation Engineering (MindTap Course List)

8th Edition

ISBN:9781305081550

Author:Braja M. Das

Publisher:Braja M. Das

Chapter6: Vertical Stress Increase In Soil

Section: Chapter Questions

Problem 6.4P: Refer to Figure P6.4. A strip load of q = 900 lb/ft2 is applied over a width B = 36 ft. Determine...

See similar textbooks

Similar questions

Note : point load = 12.289KN/m *3 = 36.867KN . 1) calculate the reaction forces 2) Calculate and draw the Bending moment digram, and what is the value of maximium bending moment ? 3) calculate and draw the shear force digram and what is the value of the maximum shear force ?
Compute the value of EIy at 2.5 m from the left support.
If a twisting moment of 1100 N-m is impressed upon a 4.4 cm ø shaft;1. What is the maximum shearing stress (in MPa) developed? 2. What is the angle of twist (in rad) in a 150 cm length of the shaft?Gsteel = 85 GPa.
Application of Equilibrium Solve for the value of unknown variables Given : P1 = 10KN W1 = 5 KN/m P2 = 6 KN W2 = 10 KN/m θ= 58° a.Unknowns: RAy , RBy , RBx , MB b. compute the internal shear 1.35 m to the right of support A
3.) For the given couple-force system shown, determine the horizontal component of Q. P= 335 KN Q= 44 KN R=74 KN-m Z=263 KN
A flat belt connects pulley A to pulley B. The coefficients of friction are ms = 0.25 and mk = 0.20 between both pulleys and the belt. Knowing that the maximum allowable tension in the belt is 600 N, determine the largest torque (in N.m) which can be exerted by the belt on pulley A.
The 600mm diameter steel shaft is subjected to the torques shown. Determine torsion at each section. Assume G = 75 GPa. P=74 1.5P= 1.5(74) = 111 kNm 3P= 3(74)= 222kNm
Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.15 and μB = 0.25. Determine the angle θ which will cause motion of one ofthe blocks. What is the friction force under each of the blocks when this occurs? The spring has a stiffness of k = 2 lb/ft and is originally unstretched.
The compound shaft shown consists of a solid shaft AB made ofbronze [G = 40 GPa], and a solid shaft BC made of steel [G = 85 GPa].The compound shaft is subjected to torques TB = 700 N.m at point B,and TC = 200 N.m at point C, acting in opposite directions.[LAB = 900 mm, LBC = 1200 mm, dAB = 35 mm, dBC = 25 mm]Determine (a) the internal torques in shafts AB and BC, (b) the shearstresses in shafts AB and BC, and (c) the angle of twist at point C.
The rectangular gate has dimensions: h=.8m; b=2.4m;c=0.4m;weighs 2 tons. It slides on a plane of inclination θ=70° and serves to plug the conduit of a dam. Considering that the coefficient of friction is 0.35. Determine the magnitude of the tension T needed to move the gate when the water level reaches the elevation of 62 m.
A 5-m STEEL SHAFT ROTATING AT 2Hz HAS 70 KW APPLIED AT GEAR THAT IS 2m FROM THE LEFT END WHERE 20 KW ARE REMOVED.AT THE RIGHT END, 30KW ARE REMOVED AND ANOTHER 20 KW LEAVES THE SHAFT AT 1.5m FROM THE RIGHT END. (A) FIND THE UNIFORM SHAFT DIAMETER SO THAT THE SHEARING STRESS WILL NOT EXCEED 60 MPa. (B) IF A UNIFORM SHAFT DIAMETER OF 100mm IS SPECIFIED, DETERMINE THE ANGLE BY WHICH ONE OF THE SHAFTS LAGS BEHIND THE OTHER END. USE G = 83 GPa.
Write legibly and box the final answer. Determine the maximum mass (kg) that can be lifted at E if the 11.670-HP motor A has a maximum rotation speed of 301.20 rpm where the diameter of the shaft AB and CE are 10.237 mm and 15.402 mm, respectively. The allowable shearing stress is 1.001 GPa with F.S. of 1.76 for both shafts. Neglect the deformation of the cable at E. Where: radius of the gear at B, rB = 113.10 mm radius of the gear at C, rc = 158.58 mm